Let $\mathbf{u} := (u_1, u_2, u_3)^T$ and $\mathbf{v} := (v_1, v_2, v_3)^T$ be column vectors, and the Nabla operator $\nabla := (\partial_1, \partial_2, \partial_3)^T$ as a column vector. Under the basic rules of matrix multiplication, it's easy to find the inner product $\mathbf{u} \cdot \mathbf{v} = \mathbf{u}^T \mathbf{v}$, and the dyadic product $\mathbf{u} \otimes \mathbf{v} = \mathbf{u} \mathbf{v}^T$. Then the divergence can be shown as $\nabla \cdot \mathbf{u} = \nabla^T \mathbf{u} = \partial_1 u_1 + \partial_2 u_2 + \partial_3 u_3$, and we have \begin{align} \nabla \otimes \mathbf{u} = \nabla \mathbf{u}^T = \begin{bmatrix} \partial_1 u_1 & \partial_1 u_2 & \partial_1 u_3 \\ \partial_2 u_1 & \partial_2 u_2 & \partial_2 u_3 \\ \partial_3 u_1 & \partial_3 u_2 & \partial_3 u_3 \end{bmatrix} \end{align} which is exactly the Jacobian of $\mathbf{u}$, denoted by $D(\mathbf{u})$.

On the other hand, we also have the gradient of a vector denoted by $\nabla \mathbf{u}$, which is by definition the transpose of the Jacobian \begin{align} \nabla \mathbf{u} := \begin{bmatrix} \partial_1 u_1 & \partial_2 u_1 & \partial_3 u_1 \\ \partial_1 u_2 & \partial_2 u_2 & \partial_3 u_2 \\ \partial_1 u_3 & \partial_2 u_3 & \partial_3 u_3 \end{bmatrix} \end{align}

So, is there an identity that $\nabla \mathbf{u} = (\nabla \otimes \mathbf{u})^T = \mathbf{u}^T \nabla$? [Q1]

In addition, with interior and exterior derivatives I find the following identity \begin{align} \nabla \mathbf{u} = \nabla \cdot \mathbf{u} + \nabla \wedge \mathbf{u} \end{align}

However, if $\mathbf{u}$ is a regular vector field then it is unreasonable, because the LHS is a (Jacobian$^T$) matrix and the RHS is a scalar plus a vector (how can they be added?). [Q2]

Please answer my questions with using mathematical knowledge not beyond undergraduate level. Thanks!

  • $\begingroup$ Regarding Q1, the identity you mention is correct as long as you manipulate the quantities in them appropriately, keeping in mind that $\nabla$ isn't -really- a vector but a vector-like object. You don't see many identities like this because they lend themselves to misinterpreting $\nabla$ as an actual vector even though they're right component-wise. No idea about Q2 though. $\endgroup$ – aghostinthefigures Jun 12 '18 at 19:20

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